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dexar [7]
2 years ago
3

The number 6 appears above the letter C in one box of the periodic table. The number 5 appears in the box to the left of the box

labeled C. The number 7 appears in the box to the right of the box labeled C. Which of the following does the number 6 tell you about the element described in the box labeled C?
Chemistry
1 answer:
Scorpion4ik [409]2 years ago
7 0

Answer: D

Explanation The average atomic mass of C

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In Part A, you found the amount of product (1.80 mol P2O5 ) formed from the given amount of phosphorus and excess oxygen. In Par
Finger [1]

First step is to balance the reaction equation. Hence we get P4 + 5 O2 => 2 P2O5

Second, we calculate the amounts we start with

P4: 112 g = 112 g/ 124 g/mol – 0.903 mol

O2: 112 g = 112 g / 32 g/mol = 3.5 mol

Lastly, we calculate the amount of P2O5 produced.

2.5 mol of O2 will react with 0.7 mol of P2O5 to produce 1.4 mol of P2O5.

This is 1.4 * (31*2 + 16*5) = 198.8 g

3 0
2 years ago
A blue circle labeled proton and A overlaps a pink circle labeled electron and C. The overlap is labeled B. The Venn diagram com
Sedaia [141]

Answer: The answer would be, "Has mass of 1 amu".

Explanation:

8 0
2 years ago
Read 2 more answers
Acetaldehyde shows two UV bands, one with a lmax of 289 nm ( 5 12) and one with a lmax of 182 nm ( 5 10,000). Which is the n p*
UkoKoshka [18]

Answer:

The bands are due to:

λmax = 289 nm n→π* transition (E = 12)

λmax = 182 nm π→π* transition (E=10000)

Explanation:

The two types of acetaldehyde transition are as follows:

n→π* and π→π*

From the attached diagram we have to:

ΔEn→π* < ΔEπ→π*

ΔEα(1/λ)

Thus:

λn→π* > λπ→π*

In n→π* spin forbidden, the intensity is low. Thus, the molar extinction E for n→π* is very low.

The same way, for π→π* spin allowed the intensity is high. Thus, the molar extinction coefficient E for π→π* is high too.

The bands are due to:

λmax = 289 nm n→π* transition (E = 12)

λmax = 182 nm π→π* transition (E=10000)

5 0
2 years ago
Three 5-l flasks, fixed with pressure gauges and small valves, each contains 4 g of gas at 273 k. flask a contains h2, flask b c
Varvara68 [4.7K]
First, please check the missing part in your question in the attachment.
a) So first, the Rank of pressure:
according to this formula PV = nRT and when n = m/Mw
PV = m/Mw * R*T
when we have the same mass m and the same V volume so P will proportional with the mole weight M as when the M is smaller the pressure will be greater 
when Mw of H2(A) = 2 g / Mw of He (B) = 4 g and Mw of CH4(C) = 16 g
∴ Pressure :
 (A) > (B) > (c)

B) The rank of average molecular kinetic energy:
when K = 3/2 KB T
when K is the average kinetic energy per molecule of gas 
and KB is Boltzmann's constant
and T is the temperature (K)
So from this equation, we can know that K only depends on T value, and when we have the T constant here for A, B, and C So the rank of K will be like the following:
∴ A = B = C
C) the rank of diffusion rate after the valve is opened:
according to this formula:
R2/R1 = √M1/M2
from this equation, we can see that diffusion is proportional to the reciprocal of the molecular mass M so,
when Mw H2 (A) = 2 g & Mw He(B) = 4 g & CH4 (C) = 16 g
∴ the rank of diffusion:
A > B > C

D) The rank of the Total kinetic energy of the molecules:
when we have the Mw different so it will make the no.of molecules differs as when the Mw is low the no.of molecules will be hight, and when the average molecular kinetic energy equals. so the total kinetic energy will depend on no. of molecules 
∵ Mw A < Mw B < Mw C 
∴no .of molecules of A > B >C
∴ the rank of total kinetic energy is:
A > B > C

e) the rank of density:

when ρ = m/ v 
and m is the mass & v is the volume and we have both is the same for A, B, and C
so the density also will be the same, ∴ the rank of the density is:
A = B = C

F) the rank of the collision frequency:
as the no.of molecules increase the collision frequency increase and depend also on the velocity and it's here the same.
∴ Collision frequency will only depend on the no.of molecules
we have no.of molecules of A > B > C as Mw A < B < C 
∴the rank of the collision frequency is:
A > B > C 

 



7 0
2 years ago
A container of gas is heated from 250 K to 303 K. What is the new pressure if the initial pressure is 880 kPa? (Assume constant
evablogger [386]

Answer:

The new pressure at constant volume is 1066.56 kPa

Explanation:

Assuming constant volume, the pressure is diectly proportional to the temperature of a gas.

Mathematically, P1/T1 = P2 /T2

P1 = 880 kPA= 880 *10^3 Pa

T1 = 250 K

T2 = 303 K

P2 =?

Substituting for P2

P2 = P1 T2/ T1

P2 = 880 kPa * 303 / 250

P2 = 266,640 kPa/ 250

P2 = 1066.56 kPa.

The new pressure of the gas is 1066.56 kPa

5 0
2 years ago
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